= Solution
In sinusoidal <steady state>, <complex impedance> is the <voltage> <phasor> divided by the current <phasor>. Its <modulus> is the <amplitude> ratio and its <argument> is the <voltage> lead. With the $e^{i\omega t}$ convention, $I=dq/dt=i\omega CV$, so \b[$Z_C=1/(i\omega C)$].
The PDF labels $V_1$ across $R_1$, not across the complete driven network. If $I$ is the current through $R_1$, then $V_1=IR_1$ and $I=Y_pV_2$, where
$$
Y_p=\frac1R+i\left(\omega C-\frac1{\omega L}\right).
$$
The <resistor-referenced parallel-resonator response> is
$$
\boxed{H(\omega)=\frac{V_2}{V_1}=\frac{R/R_1}{1+iR(\omega C-1/(\omega L))}.}
$$
Writing $b(\omega)=\omega C-1/(\omega L)$,
$$
|H|=\frac{R/R_1}{\sqrt{1+R^2b^2}},\qquad
\arg H=-\tan^{-1}(Rb).
$$
The maximum occurs at $\omega_0=1/\sqrt{LC}$, where \b[$|H|=R/R_1$ and the <phase> is zero]. Below <resonance> the <phase> is positive, tending to $+\pi/2$; above it the <phase> is negative, tending to $-\pi/2$. The <amplitude> tends to zero at both extremes: the <inductor> shorts the parallel network at low <frequency>, and the <capacitor> does so at high <frequency>. At <resonance> their reactive currents cancel, leaving only the <resistor> current.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1-ia2012-parallel-response.png]
{title=Resistor-referenced parallel-resonator amplitude and phase, normalized without choosing circuit component values}
If instead $V_{\rm in}$ denotes the <voltage> across the complete circuit, its different transfer ratio is $V_2/V_{\rm in}=1/[1+R_1/R+iR_1b]$, with peak $R/(R+R_1)$. This distinction follows directly from the printed <voltage> endpoints.
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